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# LectureNotes16 - ECS 120 Lesson 16 Turing Machines Pt 2...

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ECS 120 Lesson 16 – Turing Machines, Pt. 2 Oliver Kreylos Friday, May 4th, 2001 In the last lesson, we looked at Turing Machines, their differences to finite state machines and pushdown automata, and their formal definition. Today we will concentrate on the computation of Turing Machines. First, let us recollect the informal notion of how a Turing Machine processes an input word: 1. Before a Turing Machine starts computation, the input word w = w 1 w 2 . . . w n Σ * will be written into the leftmost n cells of the work tape. All other tape cells will be filled with the special blank sym- bol Γ. The read/write head will be located over the leftmost tape cell, and the state control will be in the start state q 0 . 2. In each step of computation, the machine will read the character a Γ located underneath the read/write head and transition from its current state q 1 Q guided by the transition function δ : If δ ( q 1 , a ) = ( q 2 , b, D ), then the machine will transition to state q 2 Q , will replace the char- acter in the tape cell underneath the read/write head by the tape char- acter b Γ, and will move the read/write head one cell to the left if D = L or one cell to the right if D = R. If the read/write head is currently located over the leftmost tape cell, and D = L, then the read/write head will remain in its current position. 3. To finish the computation, the machine must enter either the accept state q accept Q or the reject state q reject Q . In the former case, the machine accepts the input word; in the latter case, it rejects. If the machine never enters either the accept state or the reject state, computation will continue forever and the machine will never halt. 1

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1 Formal Definition of Turing Machine Com- putation To formalize our understanding of how a Turing Machine computes, we have to find a way of describing its current configuration. In the study of PDAs, we encountered instantaneous descriptions , which exactly describe the status of computation of a PDA. We will use the same mechanism for Turing Machines as well. To specify a Turing Machine’s configuration, we have to know: The current state it is in, The current contents of the work tape, and the current position of the read/write head. As for PDAs, we will encode the work tape’s contents as a string over Γ. Since all the tape cells to the right of the rightmost position the read/write head ever was in are filled with blank symbols, we will not explicitly include those into the tape string. To encode the read/write head’s position, we will split the tape string into two parts u, v Γ * ; the first, u , containing all characters from tape cells to the left of the read/write head’s current position, and the second, v , containing the characters underneath the tape head and to the right. Together with the machine’s current state, an instantaneous description (ID) for a Turing Machine is a triple ( u, q, v ) Γ * × Q × Γ * , where the current state is q , the current tape content is uv * ( u followed by v
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LectureNotes16 - ECS 120 Lesson 16 Turing Machines Pt 2...

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